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甲基碘与[Rh(acac)(CO)(PPh3)]的氧化加成:产物立体化学和反应机理的实验和理论研究。

Methyl iodide oxidative addition to [Rh(acac)(CO)(PPh3)]: an experimental and theoretical study of the stereochemistry of the products and the reaction mechanism.

机构信息

Department of Chemistry, University of the Free State, 9300, Bloemfontein, Republic of South Africa.

出版信息

Dalton Trans. 2011 Aug 28;40(32):8226-37. doi: 10.1039/c1dt10271k. Epub 2011 Jul 15.

Abstract

Density functional theory was used to investigate the oxidative addition and subsequent carbonyl insertion and deinsertion steps of the reaction of methyl iodide to a rhodium(I) acetylacetonato complex of the formula [Rh(acac)(CO)(PPh(3))] (Hacac = acetylacetone). This process has been studied experimentally for many rhodium β-diketonato complexes, but, to the best of our knowledge, this is the first systematic computational study of the complete reaction sequence. Experimental (1)H techniques complement the theoretical results on the stereochemistry of the reaction intermediates and products. (1)H NMR also revealed the existence of a second rhodium(III)-acyl product, which has not been previously observed in this reaction. The calculated Gibbs free energy of activation of the oxidative addition reaction is 71 kJ mol(-1), which is in agreement with the experimental value of 82(1) kJ mol(-1). The DFT-calculated oxidative addition corresponds to an associative S(N)2 nucleophilic attack by the rhodium metal centre on the methyl iodide, which is in agreement with calculated and experimental (in brackets) activation parameters of the reaction, 27 (38.8) kJ mol(-1) for ΔH((≠)) and -147 (-146) J K(-1) mol(-1) for ΔS((≠)).

摘要

密度泛函理论被用于研究甲基碘与[Rh(acac)(CO)(PPh(3))](Hacac = 乙酰丙酮)的铑(I)乙酰丙酮配合物的氧化加成以及随后的羰基插入和脱插入反应步骤。该过程已经在许多铑β-二酮配合物中进行了实验研究,但据我们所知,这是对完整反应序列的首次系统计算研究。实验(1)H 技术补充了关于反应中间体和产物立体化学的理论结果。(1)H NMR 还揭示了存在第二种铑(III)-酰基产物,这在该反应中以前没有观察到。氧化加成反应的计算吉布斯自由能活化为 71 kJ mol(-1),与实验值 82(1) kJ mol(-1)一致。DFT 计算的氧化加成对应于铑金属中心对甲基碘的亲核进攻的 S(N)2 缔合,这与反应的计算和实验(括号内)活化参数一致,27(38.8)kJ mol(-1)为 ΔH((≠)),-147(-146)J K(-1) mol(-1)为 ΔS((≠))。

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